sql面试题，求大神解答

[菜鸟军爷]

[凯歌6]

一: Select a.name,a.kemu,a.fenshu,(Select avg(b.fenshu) From t_grade b where a.kemu=b.kemu GROUP BY b.kemu ) From t_grade a Where a.fenshu<(Select avg(b.fenshu) From t_grade b where a.kemu=b.kemu GROUP BY b.kemu ) GROUP BY a.kemu 二： SELECT name,kemu,fenshu from t_grade where name not in(select name from t_grade where fenshu<=80) 三: SELECT name as 姓名, MAX( CASE WHEN kemu='语文' THEN fenshu END ) AS 语文, MAX( CASE WHEN kemu='数学' THEN fenshu END ) AS 数学 FROM t_grade GROUP BY name 四： select name,kemu, case when fenshu>=90 then '优秀' when fenshu<90 and fenshu>=80 then '中等' when fenshu<80 then '合格' end as fenshu from t_grade

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[qq_16680181]

select a.*,b.avg_fenshu FROM chengji a LEFT JOIN (select kemu,AVG(fenshu) as avg_fenshu FROM chengji GROUP BY kemu) b on a.kemu = b.kemu where a.fenshu < b.avg_fenshu select name,GROUP_CONCAT(fenshu) from chengji where fenshu > 80 GROUP BY NAME HAVING count(*) > 1 SELECT a. NAME, substring_index(a.fenshus, ',', 1) AS '数学', substring_index(a.fenshus, ',', - 1) AS '语文' FROM ( SELECT NAME, GROUP_CONCAT(kemu ORDER BY kemu) AS kemus, GROUP_CONCAT(fenshu ORDER BY kemu) AS fenshus FROM chengji GROUP BY NAME ) a SELECT name,kemu,if(fenshu < 80,'合格',if(fenshu >= 90,'优秀','中等')) FROM chengji

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